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AAS, HL, and Invalid Tests: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Turn the third angle into usable information

Triangles ABC and DEF have the following givens:
$\begin{aligned} m\angle A&=56^\circ \\ m\angle D&=56^\circ \\ m\angle B&=61^\circ \\ m\angle E&=61^\circ \\ AC&=24\text{ cm} \\ DF&=24\text{ cm}.\end{aligned}$
Find the third angle in each triangle. Identify a congruence criterion from the original givens, then explain how ASA can establish the same result.

Text description of this figure

The given angles at A and D are 56 degrees, and those at B and E are 61 degrees. AC and DF are the supplied equal sides of 24 centimeters. The third angles have no values or equality marks before the calculation.
1. Hint 1
  The known side AC is not between the given angles at A and B.
2. Hint 2
  Find C and F from the angle sum, then use the endpoint angles of AC and DF.
Answer

Angles C and F each measure $63^\circ$ . AAS gives $\triangle ABC\cong\triangle DEF$ ; alternatively use angles A,C and D,F with the included sides AC,DF for ASA.

Full solution

The third angle in each triangle is

$180^\circ-56^\circ-61^\circ=63^\circ.$

The given side AC is nonincluded between angles A and B, and DF occupies the corresponding position.
Thus the original information gives AAS.

The now-known angles C and F match.
Use angles A and C with side AC, and angles D and F with side DF.
The side is included between these angle pairs, so ASA proves the same congruence.

Answer

Angles C and F each measure $63^\circ$ . AAS gives $\triangle ABC\cong\triangle DEF$ ; alternatively use angles A,C and D,F with the included sides AC,DF for ASA.

Key idea

Finding the third angle lets you choose the pair that includes the known side.
Problem 2 Read the right angles before naming the match

Triangle KLM is right at L. Triangle PQR is right at Q. The given side equalities are
$\begin{aligned}KM&=PR.\\LM&=PQ.\end{aligned}$
Decide whether congruence follows and write the triangle corresponding to KLM in order. State the side corresponding to KL.

Text description of this figure

Triangles KLM and PQR have right-angle squares at L and Q. KM and PR, the given hypotenuses, have two ticks. The supplied equal legs LM and PQ have one tick. No match is marked on KL or RQ.
1. Hint 1
  The right-angle vertices must correspond, and the sides opposite them are the hypotenuses.
2. Hint 2
  LM matches PQ; with L matched to Q, determine where M and K must go.
Answer

$\triangle KLM\cong\triangle RQP$ by HL; KL corresponds to RQ.

Full solution

The hypotenuses KM and PR have equal lengths.
LM and PQ are equal legs, and both triangles are right, so HL applies.

The right angles match L to Q.
The equal leg then matches M to P, leaving K matched to R.
Therefore

$\begin{aligned}\triangle KLM&\cong\triangle RQP.\end{aligned}$

Side KL corresponds to RQ.

Answer

$\triangle KLM\cong\triangle RQP$ by HL; KL corresponds to RQ.

Key idea

The right-angle vertex and one leg pair fix the ordered correspondence in an HL comparison.
Problem 3 Separate insufficiency from noncongruence

For each record, state whether it proves the triangles congruent, proves them noncongruent, or leaves congruence undecided.

Record A. All three corresponding angle pairs match, and one corresponding side pair has lengths 8 centimeters and 13 centimeters.

Record B. Two corresponding side pairs and a corresponding nonincluded angle pair match. No additional measurements or right-angle information are supplied.

Text description of this figure

The figure represents Record A only. Three corresponding angle pairs have matching one-, two-, and three-arc patterns. The supplied corresponding sides are labeled 8 centimeters and 13 centimeters. No other lengths are supplied, and Record B is not pictured.
1. Hint 1
  For Record A, ask which side-length condition a congruence would require.
2. Hint 2
  In Record B, a failed general criterion does not prove that the particular triangles fail to match.
Answer

Record A proves noncongruence. Record B leaves congruence undecided.

Full solution

Record A.
Congruent triangles must have equal corresponding side lengths.
The recorded lengths are unequal:

$8\ne13.$

These triangles cannot be congruent despite the matching angles.

Record B.
The record supplies unrestricted SSA.
The two-intersection construction shows why these data alone need not fix the triangle.
A congruent pair could also satisfy this record, so neither congruence nor noncongruence follows.

Answer

Record A proves noncongruence. Record B leaves congruence undecided.

Key idea

An unequal corresponding length proves noncongruence; an insufficient criterion only leaves the question open.
Problem 4 A side equality earned by AAS

Triangles UVW and XYZ satisfy
$\begin{aligned} \angle U&\cong\angle X \\ \angle W&\cong\angle Z \\ UV&=XY.\end{aligned}$
The corresponding sides have lengths
$\begin{aligned} UW&=8n+2\text{ cm} \\ XZ&=5n+26\text{ cm}.\end{aligned}$
Find n and their common length.

Text description of this figure

Angles U and X have one arc, and W and Z have two. The given sides UV and XY have matching ticks. UW and XZ carry their variable expressions without equality ticks before congruence has been proved.
1. Hint 1
  UV and XY are corresponding nonincluded sides for the given angle pairs.
2. Hint 2
  After proving congruence, set the corresponding length expressions equal.
Answer

$n=8$ and both lengths are 66 centimeters.

Full solution

The two equal-angle pairs and the corresponding nonincluded side pair establish congruence by AAS:

$\begin{aligned}\triangle UVW&\cong\triangle XYZ.\end{aligned}$

Corresponding sides UW and XZ therefore satisfy

$\begin{aligned}8n+2&=5n+26.\end{aligned}$

Subtract 5n and then 2 to obtain

$\begin{aligned}3n&=24.\end{aligned}$

Divide by 3:

$\begin{aligned}n&=8.\end{aligned}$

Substitute in each expression:

$\begin{aligned}8(8)+2&=66.\\5(8)+26&=66.\end{aligned}$

Both lengths are 66 centimeters.

Answer

$n=8$ and both lengths are 66 centimeters.

Key idea

A length equation needs a justified correspondence before it is used.
Problem 5 Match a right triangular brace

An ideal rigid brace is a right triangle with a hypotenuse of 29 centimeters and one leg of 20 centimeters. A right triangular template has a hypotenuse of 29 centimeters and legs of 20 and 21 centimeters. Decide whether the brace must fit the template after rigid motions, and find its other leg. Treat all lengths and right angles as exact.
1. Hint 1
  Compare a hypotenuse with a hypotenuse and the 20-centimeter leg with its match.
2. Hint 2
  The remaining leg follows from congruence; no separate length formula is needed.
Answer

The brace fits the template by HL; its other leg is 21 centimeters.

Full solution

Both triangles are right.
Their hypotenuses match at 29 centimeters and one corresponding leg pair matches at 20 centimeters.
HL therefore proves congruence.

The other leg corresponds to the template side of length 21 centimeters, so it also measures 21 centimeters.
A rigid-motion placement aligns every part of the brace with the template.

Answer

The brace fits the template by HL; its other leg is 21 centimeters.

Key idea

A known comparison triangle can supply an inaccessible length once congruence is established.
Problem 6 Two legs instead of a hypotenuse

Triangles ABC and DEF are right at A and D.
$\begin{aligned} AB&=16\text{ cm} \\ DE&=16\text{ cm} \\ AC&=25\text{ cm} \\ DF&=25\text{ cm}.\end{aligned}$
Establish congruence using the stated side pairs directly. Explain why naming this comparison HL would misidentify the givens.

Text description of this figure

Triangles ABC and DEF have right-angle squares at A and D. Their given corresponding legs are marked and labeled 16 centimeters and 25 centimeters. Their slanted sides are unmarked and unmeasured.
1. Hint 1
  The given sides meet at the right angles.
2. Hint 2
  Which side is opposite the right angle in each triangle?
Answer

$\triangle ABC\cong\triangle DEF$ by SAS. The equal given sides are two leg pairs, not a hypotenuse and a leg.

Full solution

The side pairs AB,DE and AC,DF surround the equal right angles at A,D.
SAS therefore gives

$\triangle ABC\cong\triangle DEF.$

The hypotenuses are BC and EF, which are not among the stated side equalities.
HL needs an equal hypotenuse pair as well as an equal leg pair.
The direct criterion here is SAS on the two given legs and their included right angle.

Answer

$\triangle ABC\cong\triangle DEF$ by SAS. The equal given sides are two leg pairs, not a hypotenuse and a leg.

Key idea

Choose the criterion from the roles of the supplied parts, not just from the presence of right triangles.
Problem 7 Test an angle-only guarantee

A claim says that any two triangles with three matching corresponding angle pairs must be congruent. Give a concrete counterexample with specified side lengths, and verify both the matching angles and the failure of congruence.
1. Hint 1
  Choose triangles whose angles are known without calculating any side ratios.
2. Hint 2
  Choose a family whose angle measures stay fixed while its side length changes.
Answer

Example: equilateral triangles of sides 11 cm and 19 cm, each with three $60^\circ$ angles. Accept any pair with three verified matching angle pairs and an unequal corresponding side.

Full solution

Construct an equilateral triangle of side 11 centimeters and another of side 19 centimeters.
The base-angle theorem makes each equiangular, and its angle sum makes every angle $60^\circ$ .

Thus all three corresponding angle pairs match.
But every side in one triangle has length 11 while every side in the other has length 19, and

$11\ne19.$

A rigid motion cannot match them.

Answer

Example: equilateral triangles of sides 11 cm and 19 cm, each with three $60^\circ$ angles. Accept any pair with three verified matching angle pairs and an unequal corresponding side.

Key idea

One fully specified counterexample defeats a universal congruence claim.
Problem 8 Recover the HL argument

In the shown configuration, triangles ABC and ABF are right at A, and C and F are on opposite sides of AB. The given hypotenuse equality is
$\begin{aligned}BC&=BF.\end{aligned}$
Prove that triangles ABC and ABF are congruent without using HL itself or assuming $AC=AF$ .

Text description of this figure

Triangles ABC and ABF share leg AB and have right angles at A. C is above line AB and F is below it. Their hypotenuses BC and BF have matching ticks; no equality of AC and AF is given.
1. Hint 1
  First explain why C, A, and F lie on one line, with A between C and F.
2. Hint 2
  The equal hypotenuses make triangle BCF isosceles. Use its base angles, then the equal right angles in the smaller triangles.
Answer

$\triangle ABC\cong\triangle ABF$ by the isosceles base-angle theorem, the third-angle theorem, and SAS.

Full solution

The right-angle givens put AC and AF on the unique perpendicular to AB through A.
Since C and F are on opposite sides, C,A,F are collinear with A between them.

In triangle BCF, $BC=BF$ , so the base angles at C and F are congruent.
These are the smaller triangles' angles BCA and BFA.

Both smaller triangles are also right at A, so the third-angle theorem makes angles ABC and ABF congruent.

Now AB is shared.
The hypotenuse equality is

$\begin{aligned}BC&=BF.\end{aligned}$

The equal angles at B are included between these side pairs.
SAS gives

$\begin{aligned}\triangle ABC&\cong\triangle ABF.\end{aligned}$

Answer

$\triangle ABC\cong\triangle ABF$ by the isosceles base-angle theorem, the third-angle theorem, and SAS.

Key idea

HL can be justified by an isosceles triangle and SAS without calculating the missing legs.
Problem 9 Translate a criterion into a construction

Construct triangle HJK on paper with
$\begin{aligned} m\angle H&=48^\circ \\ m\angle J&=79^\circ \\ HK&=13\text{ cm}.\end{aligned}$
Use a ruler and protractor for the numerical drawing. Describe how to place the unknown vertex from these data, and explain why every exact successful drawing is congruent.
1. Hint 1
  Find angle K before drawing the two endpoint rays of the known side HK.
2. Hint 2
  The rays from H and K meet at J. The known side is included between those endpoint angles.
Answer

Draw HK and rays into the same half-plane, meeting at J:
$\begin{aligned}HK&=13\text{ cm}\\m\angle H&=48^\circ\\m\angle K&=53^\circ.\end{aligned}$
ASA fixes every exact result; the original data are AAS.

Full solution

The third angle is

$180^\circ-48^\circ-79^\circ=53^\circ.$

Draw HK with length 13 centimeters.
On one side of its line, draw a ray from H making 48 degrees with HK and a ray from K making 53 degrees with KH.
Name their intersection J.

The angle at J is then

$180^\circ-48^\circ-53^\circ=79^\circ.$

This meets the original requirement.

Every exact successful drawing has matching angles at H and K and their included side HK, so ASA proves congruence.
A drawing on the other side is a reflected congruent copy.
The original known angles H,J and nonincluded side HK form AAS data.

Answer

Draw HK and rays into the same half-plane, meeting at J:
$\begin{aligned}HK&=13\text{ cm}\\m\angle H&=48^\circ\\m\angle K&=53^\circ.\end{aligned}$
ASA fixes every exact result; the original data are AAS.

Key idea

AAS construction becomes an ASA construction once the third angle is found.
Problem 10 Keep both compass intersections

On paper, draw a horizontal ray from A. Draw a second ray 30 degrees above it and mark C at distance 12 centimeters from A. Draw a circle centered at C with radius 9 centimeters, keeping the compass arc that crosses the horizontal ray.

Locate every intersection on that horizontal ray, join each one to C, and compare the resulting triangles. State the two given side lengths and the given angle they share. Explain whether all of the triangles are congruent. You may use a scaled drawing; state its scale and keep the same scale for both triangles. No exact calculation of the other side lengths is required.
1. Hint 1
  The circle may cross the ray at more than one place; do not stop after the first intersection.
2. Hint 2
  A circle radius supplies the side from C. Different intersection positions on one ray give different distances from A.
Answer

There are two intersections beyond A, giving noncongruent triangles. Their shared givens are
$\begin{aligned}AC&=12\text{ cm}.\\CB&=9\text{ cm}.\\m\angle A&=30^\circ.\end{aligned}$
Their AB lengths differ.

Full solution

For example, use one drawn centimeter for every two centimeters of the intended triangle.
Mark C 6 drawn centimeters from A and use a radius of 4.5 drawn centimeters.
The two intersections correspond to points B on the original ray.

Name the nearer one B₁ and the farther one B₂.
Both triangles share

$AC=12\text{ cm}.$

Since both intersection points are on the circle centered at C, each has

$CB=9\text{ cm}.$

Both use the same 30-degree angle at A.

The distinct intersection points are at different distances from A along the same ray, so

$AB_1\ne AB_2.$

Each triangle has side lengths 12 and 9, but their third side lengths differ.
Their full side-length lists cannot match under any vertex order, so the triangles are noncongruent.

This one fixed data set supplies a counterexample to unrestricted SSA being a congruence criterion.
It does not imply that every SSA construction has two triangles.

Answer

There are two intersections beyond A, giving noncongruent triangles. Their shared givens are
$\begin{aligned}AC&=12\text{ cm}.\\CB&=9\text{ cm}.\\m\angle A&=30^\circ.\end{aligned}$
Their AB lengths differ.

Key idea

Keep all intersections in a construction: a second valid placement can disprove a proposed uniqueness rule.

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